---
title: "Black and grey body emission numerical"
subject: "Heat Transfer"
chapter: 4
key: "semester_5/heat_transfer:blackbody-wien-emissive-power-numerical"
source: https://www.bctengineer.com/q/semester_5/heat_transfer/blackbody-wien-emissive-power-numerical
site: BCT Engineering
---

# Black and grey body emission numerical

Heat Transfer · Chapter 4: Radiation Heat Transfer

- Asked in: year not given, 6 marks
- Video: [Blackbody Radiation Numericals | Emissive Power & Rate of Heat Loss | Stefan’s Law Explained](https://www.youtube.com/watch?v=rX7LajSdOLY) — Understanding.Concepts.Of.Physics
- Answer checked by Claude Sonnet 5.5 on 2026-10-14
- Page: https://www.bctengineer.com/q/semester_5/heat_transfer/blackbody-wien-emissive-power-numerical
- Whole chapter: https://www.bctengineer.com/md/semester_5/heat_transfer/chapter-4

## Question

A tungsten filament furnace surface is at 1500 K. Calculate (a) the wavelength at which the monochromatic emissive power is maximum, (b) the total emissive power if it behaves as a black body, and (c) the heat radiated per second from a 0.5 m^2 surface if it is a grey body with emissivity 0.8. Compare the peak wavelength with that of the sun, taken as a black body at 5800 K.

## Short answer

λ_max = 2898/1500 = 1.93 μm; E_b = σT^4 = 5.67e-8 × 1500^4 = 287 kW/m^2; for grey body ε=0.8, Q = ε E_b A = 0.8 × 287e3 × 0.5 = 114.8 kW. Sun at 5800 K peaks at 0.50 μm.

## Answer

#### (a) Peak wavelength (Wien's law)

$$
\lambda_{max} = \frac{2898}{T} = \frac{2898}{1500} = 1.932\ \mu\text{m}
$$

This is in the near-infrared; only a small tail lies in the visible range.

#### (b) Black-body emissive power (Stefan-Boltzmann law)

$$
E_b = \sigma T^4 = 5.67\times10^{-8}\times(1500)^4 = 2.870\times10^{5}\ \text{W/m}^2
$$

#### (c) Grey body radiation

$$
\begin{aligned}
E &= \varepsilon\sigma T^4 = 0.8\times2.870\times10^5 = 2.296\times10^5\ \text{W/m}^2\\
Q &= E\,A = 2.296\times10^5\times0.5 = 1.148\times10^{5}\ \text{W}
\end{aligned}
$$

#### Comparison with the sun

$$
\lambda_{max,sun} = \frac{2898}{5800} = 0.4997\ \mu\text{m}
$$

The sun peaks in the visible (green), at about 0.5 $\mu$m, about one quarter of the wavelength of the 1500 K surface.

**Answer:** $\lambda_{max} = 1.93\ \mu$m; $E_b = 287\ \text{kW/m}^2$; $Q = 114.8$ kW; sun peaks at $0.50\ \mu$m.

## Check yourself

The answer's key points as multiple-choice questions, with the correct option and why.

1. Which relation gives the wavelength of maximum monochromatic emissive power, and what is it for 1500 K?
   A. lambda_max = 2898/T, giving 1.93 micrometre
   B. lambda_max = 2898 x T, giving 4.35 x 10^6 micrometre
   C. lambda_max = 2898/T^4, giving about 0 micrometre
   D. lambda_max = T/2898, giving 0.52 micrometre
   Answer: A. Wien's law is lambda_max T = 2898 um K, so 2898/1500 = 1.93 um (near infrared). (Key point: Wien's displacement law)

2. How is the heat radiated by the grey surface found after the black-body emissive power is known?
   A. Q = E_b / epsilon
   B. Q = epsilon x sigma x T x A
   C. Q = epsilon x E_b x A = 0.8 x 2.87e5 x 0.5
   D. Q = E_b x A / (1 - epsilon)
   Answer: C. A grey body emits epsilon times the black-body power per unit area, multiplied by the area: about 114.8 kW. (Key point: Grey body emission)

3. How does the sun's peak wavelength (5800 K) compare with that of the 1500 K surface?
   A. Longer, in the infrared
   B. Equal, because both are black bodies
   C. About one quarter, in the visible near 0.5 micrometre
   D. About four times longer
   Answer: C. Peak wavelength is inversely proportional to T, so a hotter body peaks at a shorter wavelength: 0.50 um against 1.93 um. (Key point: Sun versus furnace)

## Related in Heat Transfer

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- [Composite furnace wall heat loss](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Acomposite-furnace-wall-interface-temperatures) — Heat Transfer, chapter 2
- [Radiation between two parallel grey plates](https://www.bctengineer.com/question.md?key=semester_5%2Fheat_transfer%3Aradiation-exchange-parallel-grey-plates) — Heat Transfer, chapter 4

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